Atom-Ecology

  • To settle this banal discussion about the defining characteristics of that golden curve experiment, that experiment has periods where no changes whatsoever are made in the operating protocal, and the golden difference changes dramatically. That frequently changing behaviour puts an end to all this silly talk about the validity and meaning of the curve(s) with regard to heat after power off.


    The observation of changing characteristics in the golden curve experiment does not put an end to my objections unless the system at other times exhibits nonlinear behaviour such as thresholds and nonexponential relaxations during cooling.


    You are going to have to engage with well-founded objections such as mine sooner or later. I suppose the timing is up to you however. In the meantime, perhaps Alan Smith can convince you to lay off insults such as `anon troll drool`.


  • So let's be more rigorous, Wcool = k1(T - T1), Wreact=k2(T-T2), T2 activation temperature,

    then Wreact-Wcool = (k2-k1)(T-(T2k2-T1k1)/(k2-k1)) = (k2-k1)(T - T0),

    say that k2 = k and k1 = Ak (A > 1 if k1 > k2)) lead to the everall cooling to be

    k(A-1)(T -T0), T0=(T2 -T1A)/(1-A), now assume

    A>1, T0 < T2

    <=> (T2-T1A) > T2(1-A)

    <=> (T2-T1)A > 0

    <=> T2 > T1


    which probably is the case, hence we know that under the condition A > 1, the system will decay

    with rate k1-k2 towards T2 and then the reaction deactivates and the system continous to

    cool to T1 with decay rate k1. This is true for all cases here k1 > k2.

  • But over a period of several hours we can suppose that this variable is quasi invariant


    That would be a mistake. I have just been looking at some very strange data, with suggests there is a 15 second 'bursty heat' cycle, and at times another 5 second cycle. Puzzles the hell out of me.


    As for Russ, that comment was not personal, he gets pretty steamed up by anonymity though.

  • ... the system will decay with rate k1-k2 towards T2 and then the reaction deactivates and the system continous to cool to T1 with decay rate k1.


    This is incorrect as far as I can see. In your model the temperature doesn't decline to T2 at rate k1-k2 and then decline further to T1 at a different rate. Instead it declines globally towards T0 at rate k1-k2. And T0 is a temperature that is lower than the ambient temperature in the lab (which is T1).


    Why does your system head towards a temperature lower than the lab surroundings? It is because you have assumed that the lenr effect produces cooling power below T2 and I don't see how that is a physical assumption. You have gotten around this in words by saying that the reaction "deactivates". Well that is exactly my point. There is no deactivation in your model but there is in mine. And that region of temperatures over which the lenr process deactivates is exactly the nonlinearity that introduces thresholds and nonexponential relaxations.

  • I already said we repeated it and got different results. It will be repeated again and again I expect. Your argument is a bit like a 'cant win' scenario, repeat a test and get the same result is a bad sign you suggest, personally I think that repeatable experiments are better than those with wildly different results

    It depends on how close the two curves are. If they fall exactly on top of one another, I would suspect that is generated some sort of instrument error, or method error. I have seen many results like that. Wildly different results are also bad.

  • This is incorrect as far as I can see. In your model the temperature doesn't decline to T2 at rate k1-k2 and then decline further to T1 at a different rate. Instead it declines globally towards T0 at rate k1-k2. And T0 is a temperature that is lower than the ambient temperature in the lab (which is T1).


    Why does your system head towards a temperature lower than the lab surroundings? It is because you have assumed that the lenr effect produces cooling power below T2 and I don't see how that is a physical assumption. You have gotten around this in words by saying that the reaction "deactivates". Well that is exactly my point. There is no deactivation in your model but there is in mine. And that region of temperatures over which the lenr process deactivates is exactly the nonlinearity that introduces thresholds and nonexponential relaxations.

    your wrong, I modeled the T2 as the activation temperature and T2 > T0 so it will reach T2

    and then become deactivated and no longer active e.g. no more heat generated and what remains is the

    cooling only

  • That would be a mistake. I have just been looking at some very strange data, with suggests there is a 15 second 'bursty heat' cycle, and at times another 5 second cycle. Puzzles the hell out of me.


    From your description of this strange behaviour I get the impression of a burst of heat excursions followed by a period of quiescence followed by another burst and so on with the bursts and possibly the quiescent periods having variable duration. If so then one explanation is a slow variable whose level builds up a little bit, but not much, during each heat excursion. A burst of such excursions then changes the variable sufficiently that the burst ends. Thereafter there is a decay of the slow variable until it hits levels low enough for the heat excursions to begin again. From the standpoint of the mechanism underlying individual heat excursions, the slow variable appears as a slowly changing parameter.


    Many conceptual tools have been developed over the years for thinking about the behaviour of nonlinear systems possessing multiple interacting variables and/or interacting entities. This all comes under the rubric of "nonlinear dynamical systems theory", if that is useful information for you. The theory contains terms such as limit cycle, steady state, basin of attraction, bifurcation, singular pertubation, and synchronization.


    If you wish I can amplify a bit, with particular relevance to lenr, or I could suggest some basic references where these ideas have been applied to ecology or neuroscience. Or maybe you guys know all this already.

  • your wrong, I modeled the T2 as the activation temperature and T2 > T0 so it will reach T2

    and then become deactivated and no longer active e.g. no more heat generated and what remains is the

    cooling only

    You say Wreact=k2(T-T2). Doesn't this mean that Wreact <0 when T<T2? What does a negative power mean physically for the lenr mechanism? It is in your model but what does it mean?


    You keep saying that the lenr mechanism will deactivate at some point. This is how you avoid the negative lenr power. But that is just words, you don't have the deactivation in your model. I have it explicitly in mine and that is exactly what confers the threshold and nonexponential behaviour.

  • You say Wreact=k2(T-T2). Doesn't this mean that Wreact <0 when T<T2? What does a negative power mean physically for the lenr mechanism? It is in your model but what does it mean?


    You keep saying that the lenr mechanism will deactivate at some point. This is how you avoid the negative lenr power. But that is just words, you don't have the deactivation in your model. I have it explicitly in mine and that is exactly what confers the threshold and nonexponential behaviour.

    I presented the formula and said T2 is the activation temperature meaning Wreact = 0 for T < T2

  • OK. With Wreact = 0 for T less than some activation temperature (which for moderate temperatures qualitatively matches my model), and for k_cooling (k1) greater than k_heating (k2), I see that T should exponentially decline from some high temperature down to the activation temperature and then more quickly but still exponentially decline towards T1.


    How do you match all this with the red trace in Russ George`s figure.




    Where do you think T2 is?

  • say T2 is maybe 275 degree C, and T1 is 220 degree C or whatever. No idea, we must wait for proper data to say anything. What I don't understand is that the curve seam to never go under

    200 degrees.

  • That's because the PID thermostat had been set to 200C. Or maybe 230- I don't have the figures to hand.


    This would clarify things a bit for me. But the label on the figure says that the fueled reactor has "ZERO power input". This conflicts with your information doesn't it? Or am I completely misunderstanding what input power is in your system?


    Edit: Wait. Aha!! I get it! So for the last half of the figure the fueled reactor is unpowered but the control reactor is externally heated. This means that lenr heating is much more prominent than it appears from the figure and is actually pretty much fully in play right down to at least 270 degrees C or so. If the control reactor had been left to cool naturally you might would have had something like this ...


      


    This would be much more what Jed Rothwell expected to see for heat after death. And one reason Russ George`s plot is not showing the threshold effects I expected is because the entirety of the plot is way above the activation range.

  • I must confess that I have not been following your discussion with Bruce so I'll just say 'Yes'. The heating coil is re-energised when the PID's dedicated thermocouple hits the set value- or maybe a very small amount above it to allow for system hysteresis.

  • I must confess that I have not been following your discussion with Bruce so I'll just say 'Yes'. The heating coil is re-energised when the PID's dedicated thermocouple hits the set value- or maybe a very small amount above it to allow for system hysteresis.


    In stefan's usage, T1 is laboratory ambient temperature and T2 is the temperature above which lenr heating starts to kick in

  • I'll answer for Russ on this, since he probably won't post any more. Right now we have something pretty amazing going on, but we don't yet know how useful it might be. Signs are that it will be very useful, but we need to do a lot more work to make it so. We are exploring strange territory, working with relatively novel materials, there is no map, and we are not even sure how far the territory extends. So, it would be premature to discuss commercialisation as we are still experimenting and debugging our systems - not that they are very buggy, but I believe (like many) in the value of continued incremental improvements. Right now I'm putting together a 42V 137A power bus to feed all 8 reactors with exactly the same heater power- it will only be running at a small fraction of its capacity, but does represent another variable removed and another calibration chore simplified.


    Thanks for your interest, I can only assure you that we have good intentions and no fear of hard work or spending our cash on what we perceive to be important work.

  • I'll answer for Russ on this, since he probably won't post any more. Right now we have something pretty amazing going on, but we don't yet know how useful it might be. Signs are that it will be very useful, but we need to do a lot more work to make it so. We are exploring strange territory, working with relatively novel materials, there is no map, and we are not even sure how far the territory extends. So, it would be premature to discuss commercialisation as we are still experimenting and debugging our systems - not that they are very buggy, but I believe (like many) in the value of continued incremental improvements. Right now I'm putting together a 42V 137A power bus to feed all 8 reactors with exactly the same heater power- it will only be running at a small fraction of its capacity, but does represent another variable removed and another calibration chore simplified.


    Thanks for your interest, I can only assure you that we have good intentions and no fear of hard work or spending our cash on what we perceive to be important work.

    Hi Alan,


    Finally, you have a wonderful, unexpected vacation..:)

    However, in the current excitement, you probably have a same communication mode as Rossi ..:D:D

    If you really ready to share maybe with more teams around Lenrforum, you should do more work, quickly.

    I think it should be better for your health/heart , my friend ..8)8)

    From our side we continue to try the older way as NiH..so we hope also to become excited soon as you .:)


    DF